Physics 1 Homework 11
Elastic force
Last class we discussed elasticity and elastic force. Elasticity can be described as the property of an
object to return to its original shape after the force, which caused deformation of the object, is lifted.
A lot of objects around us show elastic properties. When you are knocking on the door, for example,
the door and your knuckles experience small elastic deformation. We will discuss elastic force on the
example of a coil spring. As you are stretching or squeezing a coil spring, the spring applies the
“resistance” force to your hand. The force “tries” to return the spring to its original length. This force
is called elastic force:
(1),
where x is the extension of the coil spring which means the distance that the spring has been
stretched or compressed away from the equilibrium position (see picture below). The latter is the
position where the spring would naturally come to rest. The coefficient k in the formula is called
force constant of spring constant. The constant has units of force per unit length (usually newtons per
meter). This constant depends only of the coil spring material and shape.
X
Robert Hooke (1635-1703)
Why minus in the formula? This is because the elastic force is directed against displacement. If
we squeeze the coil spring, the elastic force “tries” to stretch it back. The formula (1) is also
called the Hooke’s law. Hooke's law is named after the 17th century British physicist Robert
Hooke.
He first stated this law in 1676 as an anagram, then in 1678 in Latin as Ut tensio, sic vis,
which means: “As the extension, so the force”. For objects that obey Hooke's law (we will call
them elastic, such as coil springs or, say, rubber bands), the force which “tries” to return the coil
spring back to the initial length is proportional to the extension x.
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All the objects change their shape or, using other words, their “lengths” along different
directions under applied force. We will call this change as “deformation”. In the case of coil
spring the deformation is the change of the coil spring’s length x. Even in case of a very low
force (say, touch of the finger) applied to a really hard object (say, a diamond or a steel ball) the
latter slightly deforms. We cannot see this deformation, but it is finite and measurable.
The deformation is called elastic if the deformed object restores its original shape after
the force is removed. We can demonstrate elastic deformation by squeezing a rubber ball or
stretching a coil spring. If the deformation stays after the force is lifted it is called plastic. The
examples are spreading butter over bread or making something from play dough.
Each object has its deformation limit. If you exceed this limit when applying force to the
object the latter breaks. Even if the object is pretty hard (which means its force constant k is
high) it may have low deformation limit. We call such objects brittle and the corresponding
quality as brittleness. The examples of brittle materials are glass and porcelain. The opposite
qualities are ductility and malleability. Ductile materials can be strongly deformed when
stretched, malleable can be strongly deformed when compressed. An example of both ductile and
malleable material is chewing gum.
Problems
1. Take a piece of any elastic thread. Make marks on the thread separated by equal distances
along the entire length of the thread (you can use a pen or marker). Then, stretch the
thread. The distances between the marks will increase. Will the distances remain equal to
each other? Try to explain the result.
2. Measure the force constant of the rubber band used for the problem 1.
3. A raw egg is very easy to break but extremely difficult to crush with one hand (if you try
to do this experiment, please, be very careful!). Is there a contradiction? How you would
characterize the material of an eggshell?
4. A load of 1kg is attached to the end of the coil spring hanging from the ceiling of an
elevator. The force constant of the spring is 100N/m. The elevator goes up with the
acceleration of the 10m/s2. Find the extension of the coil spring.
5. Solve the problem 1 for the case where the elevator goes down with the acceleration of
30m/s2. What will happen to the coil spring?
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